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May 11 The Eigenvalue Method

Homogeneous

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In all of our problems before, we have done problems with one differential equation. What if we have a series of differential equations? Here is an example.
$$\begin{matrix}x'_1 & = & 3x_1 & & & + & x_3, \\
x'_2 & = & 9x_1 & - & x_2 & + & 2x_3, \\
x'_3 & = & -9x_1 & + & 4x_2 & - & x_3\end{matrix}$$
If you need a refresher on linear algebra and matrices in general go to mathwizurd.com/linalg. Anyway, we can rewrite that like this:
$$\begin{bmatrix}x'_1 \\ x'_2 \\ x'_3\end{bmatrix} = \begin{bmatrix}3 & 0 & 1\\9 & - 1 & 2\\-9 & 4 & -1\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$$
Because the original problem is composed of constant coefficients, like what we did 3 posts before, we guess $e^{cx}$. So, for the entire system, we guess $\vec{v}e^{cx}$ If we plug it in into the general equation $x' = Ax$, we get $c\vec{v}e^{cx} = A\vec{v}e^{cx}$. We divide by $e^{cx}$ and we get $c\vec{v} = A\vec{v}$. We've seen this before: that's the definition of an eigenvalue eigenvector pair.

Once more, let's review how we got here. We started with $\vec{v}e^{cx}$. So, our solutions will be of the form: eigenvector * $ e^{\text{eigenvalue} * x}$.

Problem: Imaginary Numbers

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Continuing with the current question, we potentially have a problem. Our other eigenvalues are $-1 \pm i$. Let's start with $-1 + i$, and we'll see that we don't even need to do $-1-i$.Let's substract from the diagonal like a normal eigenvalue. We get
$$\begin{bmatrix}4 - i & 0 & 1\\9 & -i & 2\\-9 & 4 & -i\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}$$
We could solve this, or we could use a trick that the book teaches. We first let $x_1 = 1$. This is okay, because the real value of $x_1$ is any multiple of that. That means $x_3 = -4 + i$. Finally, with some arithmetic, you get $x_2 = 2-i$.

Problem: Repeat Roots

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Let's say we have:
$$\vec{x}' = \begin{bmatrix}1 & -1\\4 & -3\end{bmatrix}\vec{x}$$
The only eigenvalue is -1. The only eigenvector is $\begin{bmatrix}1\\2\end{bmatrix}$. Here we have a problem. We know one solution: $\vec{x_1} = \begin{bmatrix}1\\2\end{bmatrix}e^{-t}$. We want two equations, however. So, what can we do?
For $x_2$, we guess $c_1\left(\vec{v_1}te^{-t} + \vec{v_2}e^{-t}\right)$, where $\vec{v_1}$ is the first eigenvector. The exercise is left up to the reader to derive $$\left(A - (\lambda{}I)\right)\vec{v_2} = \vec{v_1} \text{ and } \left(A - (\lambda{}I)\right)\vec{v_1} = \vec{0} $$
It's not too hard, just write out $x'$ and solve.
As an extension, we find that $\left(A - \lambda I\right)^2\vec{v_2} = \vec{0}$.
We do that for this problem.
$$\begin{bmatrix}2 & -1 \\ 4& -2\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix} = \begin{bmatrix}1\\2\end{bmatrix}$$
$$\vec{v_2} = c_1\begin{bmatrix}1\\2\end{bmatrix} + \begin{bmatrix}0\\-1\end{bmatrix}$$ The repeated $\begin{bmatrix}1\\2\end{bmatrix}$ may be absorbed.
Our final solution becomes:
$$c_1\begin{bmatrix}1\\2\end{bmatrix}e^{-t} + c_2\left(\begin{bmatrix}1\\2\end{bmatrix}te^{-t} + \left(c_3\begin{bmatrix}1\\2\end{bmatrix} + \begin{bmatrix}0\\-1\end{bmatrix}\right)e^{-t}\right)$$
Because we get a $(c_1 + c_2*c_3)\begin{bmatrix}1\\2\end{bmatrix}e^{-t}$. Those can be combined into one constant. Our real final answer is:
$$\boxed{c_1\begin{bmatrix}1\\2\end{bmatrix}e^{-t} + c_2\left(\begin{bmatrix}1\\2\end{bmatrix}te^{-t} + \begin{bmatrix}0\\-1\end{bmatrix}e^{-t}\right)}$$